Solving equations with variables on both sides

One of the zeros of the polynomial function is 3.

Orlov [11]

Answer:

f(x)=(x−3)(x+1)(x−1)(x+2)

Step-by-step explanation:

5 0

3 years ago

Plss help meee

Alborosie

Step-by-step explanation:

Storage Space of Shed = Volume of Shed.

Volume of Shed = Volume of Rectangular level

+ Volume of Triangular roof

Volume of Rectangular Level =

Length x Width x Height

=

$12 \times 6 \times 8 \\ = 576 {ft}^{3}$

Volume of Triangular Roof =

Area of Triangular side x Length

=

$\frac{1}{2} \times base \times height \times length \\ = \frac{1}{2} \times 6 \times 4 \times 12 \\ = \frac{1}{2} \times 288 \\ = 144 {ft}^{3}$

Volume of Shed = 576 + 144

=

$720 {ft}^{3}$

3 0

3 years ago

Please answer this correctly without making mistakes

Sergeeva-Olga [200]

Answer:

$\large \boxed{\mathrm{4/5 \ cups}}$

Step-by-step explanation:

Subtract 1/10 from 9/10 to find out how much is left.

9/10 - 1/10

8/10 = 4/5

8 0

3 years ago

Read 2 more answers

A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou

sergejj [24]

Answer:

$t=\frac{2.9-3}{\frac{0.3}{\sqrt{50}}}=-2.357$

$df=n-1=50-1=49$

$p_v =P(t_{(49)}$

Since the p value is less than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true growth rate is significantly less than 3 cm per week

Step-by-step explanation:

Information given

$\bar X=2.90$ represent the sample mean for the growth

$s=0.30$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =3$ represent the value to check

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to verify if the true mean for the growth us less than 3cm per week, the system of hypothesis are:

Null hypothesis: $\mu \geq 3$

Alternative hypothesis: $\mu < 3$

We don't know the population deviation so we can use the t statistic:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{2.9-3}{\frac{0.3}{\sqrt{50}}}=-2.357$

The degrees of freedom are given by:

$df=n-1=50-1=49$

The p value for this case would be given by:

$p_v =P(t_{(49)}$

Since the p value is less than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true growth rate is significantly less than 3 cm per week

3 0

3 years ago

P(A) = 20%. P(B) = 50%. P(BIA) = P(B). What is P(A and B)?

8_murik_8 [283]

Answer:

P(A and B) = 0.1 = 10%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$